Page:Biometrika - Volume 6, Issue 1.djvu/3

Section I.

Samples of $n$ individuals are drawn out of a population distributed normally, to find an equation which shall represent the frequency of the standard deviations of these samples.

If $s$ be the standard deviation found from a sample $x_{1}x_{2}...x_{n}$ (all these being measured from the mean of the population), then

$s^{2}={\frac {S(x_{1}^{2})}{n}}-\left({\frac {S(x_{1})}{n}}\right)^{2}={\frac {S(x_{1}^{2})}{n}}-{\frac {S(x_{1}^{2})}{n^{2}}}-{\frac {2S(x_{1}x_{2})}{n^{2}}}$.

Summing for all samples and dividing by the number of samples we get the mean value of $s^{2}$ which we will write ${\bar {s}}^{2}$.

${\bar {s}}^{2}={\frac {n\mu _{2}}{n}}-{\frac {n\mu _{2}}{n^{2}}}={\frac {\mu _{2}(n-1)}{n}}$,

where $\mu _{2}$ is the second moment coefficient in the original normal distribution of $x$: since $x_{1}$, $x_{2}$, etc., are not correlated and the distribution is normal, products involving odd powers of $x_{1}$ vanish on summing, so that ${\frac {2S(x_{1}x_{2})}{n^{2}}}$ is equal to $0$.

If $M'_{R}$ represent the $R$^th moment coefficient of the distribution of $s^{2}$ about the end of the range where $s^{2}=0$,

$M'_{1}=\mu _{2}{\frac {(n-1)}{n}}$.

Again

$s^{4}$	$=\left\{{\frac {S(x_{1}^{2})}{n}}-\left({\frac {S(x_{1})}{n}}\right)^{2}\right\}^{2}$
	$=\left({\frac {S(x_{1}^{2})}{n}}\right)^{2}-{\frac {2S(x_{1}^{2})}{n}}\left({\frac {S(x_{1})}{n}}\right)^{2}+\left({\frac {S(x_{1})}{n}}\right)^{4}$
	$={\frac {S(x_{1}^{4})}{n^{2}}}+{\frac {2S(x_{1}^{2}x_{2}^{2})}{n^{2}}}-{\frac {2S(x_{1}^{4})}{n^{3}}}-{\frac {4S(x_{1}^{2}x_{2}^{2})}{n^{3}}}+{\frac {S(x_{1}^{4})}{n^{4}}}$
	$+{\frac {6S(x_{1}^{2}x_{2}^{2})}{n^{4}}}+$other terms involving odd powers of $x$, etc., which will vanish on summation.

Now $S(x_{1}^{4})$ has $n$ terms but $S(x_{1}^{2}x_{2}^{2})$ has ${\frac {1}{2}}n(n-1)$, hence summing for all samples and dividing by the number of samples we get

$M'_{2}$	$={\frac {\mu _{4}}{n}}+\mu _{2}^{2}{\frac {(n-1)}{n}}-{\frac {2\mu _{4}}{n^{2}}}-2\mu _{2}^{2}{\frac {(n-1)}{n^{2}}}+{\frac {\mu _{4}}{n^{3}}}+2\mu _{2}^{2}{\frac {(n-1)}{n^{3}}}$
	$={\frac {\mu _{4}}{n}}\{n^{2}-2n-1\}+{\frac {\mu _{2}^{2}}{n^{3}}}(n-1)\{n^{2}-2n+3\}$.

Now since the distribution of $x$ is normal, $\mu _{4}=3\mu _{2}^{2}$, hence

$M'_{2}=\mu _{2}^{2}{\frac {(n-1)}{n^{3}}}\{3n-3+n^{2}-2n+3\}=\mu _{2}^{2}{\frac {(n-1)(n+1)}{n^{2}}}$.